let $f(x)=(x^{2}-1)^{4}$. over what interval is the function decreasing?\n$(-1,1)$\n$(-1,0)$ and…

let $f(x)=(x^{2}-1)^{4}$. over what interval is the function decreasing?\n$(-1,1)$\n$(-1,0)$ and $(1,\\infty)$\n$(-\\infty,-1)$ and $(0,1)$\n$(1,\\infty)$

let $f(x)=(x^{2}-1)^{4}$. over what interval is the function decreasing?\n$(-1,1)$\n$(-1,0)$ and $(1,\\infty)$\n$(-\\infty,-1)$ and $(0,1)$\n$(1,\\infty)$

Answer

Explanation:

Step1: Find the derivative of (f(x))

Use the chain rule. If (y = u^n) where (u=x^{2}-1) and (n = 4), then (y^\prime=nu^{n - 1}\cdot u^\prime). (u^\prime=2x), so (f^\prime(x)=4(x^{2}-1)^{3}\cdot2x=8x(x^{2}-1)^{3}). Factor (x^{2}-1=(x + 1)(x - 1)), then (f^\prime(x)=8x(x + 1)^{3}(x - 1)^{3}).

Step2: Find the critical points

Set (f^\prime(x)=0). (8x(x + 1)^{3}(x - 1)^{3}=0) gives (x=-1,x = 0,x = 1).

Step3: Test intervals

  • For (x\in(-\infty,-1)), let (x=-2). Then (f^\prime(-2)=8\times(-2)\times(-2 + 1)^{3}\times(-2 - 1)^{3}=8\times(-2)\times(-1)^{3}\times(-3)^{3}=8\times(-2)\times(-1)\times(-27)=-432<0).
  • For (x\in(-1,0)), let (x =-\frac{1}{2}). Then (f^\prime(-\frac{1}{2})=8\times(-\frac{1}{2})\times(-\frac{1}{2}+1)^{3}\times(-\frac{1}{2}-1)^{3}=8\times(-\frac{1}{2})\times(\frac{1}{2})^{3}\times(-\frac{3}{2})^{3}=8\times(-\frac{1}{2})\times\frac{1}{8}\times(-\frac{27}{8})=\frac{27}{16}>0).
  • For (x\in(0,1)), let (x=\frac{1}{2}). Then (f^\prime(\frac{1}{2})=8\times\frac{1}{2}\times(\frac{1}{2}+1)^{3}\times(\frac{1}{2}-1)^{3}=8\times\frac{1}{2}\times(\frac{3}{2})^{3}\times(-\frac{1}{2})^{3}=4\times\frac{27}{8}\times(-\frac{1}{8})=-\frac{27}{16}<0).
  • For (x\in(1,\infty)), let (x = 2). Then (f^\prime(2)=8\times2\times(2 + 1)^{3}\times(2 - 1)^{3}=16\times27\times1 = 432>0).

A function (y = f(x)) is decreasing when (f^\prime(x)<0).

Answer:

((-\infty,-1)) and ((0,1)) (the third option)